Recent content by bacon

Thank you, jbunniii, for your response. The expansion of e^x certainly shows the terms I was looking for, the terms do not come from the expression, f(x)≈f(0) + f'(0)x, I thought maybe I missed something in the lecture. Thanks also for giving me more depth on GPS operations, there is quite a...

My questions are from lecture 9, MIT OCW SV Calculus, Jerison, 2009; At 27:50 he is deriving the linear approximation for the function e^(-3x)(1+x)^(-1/2)≈(1-3x)(1-1/2x)≈1-3x-1/2x+3/2x^2≈1-7/2x, for x near 0. In the last step he drops the x squared term since it is negligible(no questions so...

I will provide a summary of what's been done so far; The platform: Horizontal forces. F-F_{fric}=ma_{plat} Vertical forces. N=mg+N_{1} as I stated in post 4, this N_{1}is really just Mg. One way to think of it is that the normal force of the floor must support the weight of both the...

You are getting ahead of yourself a little bit. Parts (a) and (b) will not involve rotational issues, just Newton's Second Law. Your equations of motion for the platform look good with a subtle tweak. Specifically, N_{1}is the force of the platform on the cylinder, the force of the cylinder on...

The velocities of the two masses are not necessarily the same(most of the time they are not). Look at the equation 15tungAlbert posted, the before collision momentum(left side) is equal to the after collision momentum(right side). You are given the velocity of the 1kg mass post collision. Your...

That's ok. Remember that since momentum is conserved, the momentum before the collision is the same as that after the collision. You have correctly calculated the momentum before the collision, -2kgm/s, so that's what the total after collision momentum must be as well. Your unknown speed is the...

The question is not a bother, it's the purpose of the forum. Your analysis of the before collision momentum looks very good. The after part is almost correct. The terms to the left of the equals sign are good and on the right the first term is as well. There is a "+ 2kg", is that a typo?

I would recommend drawing a picture. Identify the forces acting on each object(i.e. the platform and cylinder), and then apply Newton's second law. F_{net}=ma=F_{1}+F_{2}+... My technique is to consider all the forces acting on an object that (a)don't "touch" it, like gravity or electrical...

The mode of vibration is the "harmonic". The second allowed harmonic is not necessarily the second harmonic. In this case, the second allowed harmonic is the "third harmonic" corresponding to n= 3. If n=2 were allowed, the second harmonic, there would be a node at the open end of the pipe, which...

Given a pipe with one open end and one closed, can you have n=2? and, more generally, can you have an even n?

I have never heard of "gravitational units", but based on your second phrase, "… she said that its the absolute unit multiplied by acceleration due to gravity…", I would guess that she is referring to what is commonly called "G's"; as in , " The astronauts experienced 4 G's during their...

Thanks for your help.

I think that I’m doing this problem correctly, but the answer seems a bit unreasonable. Can someone else check my work? A thermometer has a quartz body within which is sealed a total volume of 0.400cm^{3} of mercury. The stem contains a cylindrical hole with a bore diameter of 0.10mm. How far...

I'll take the easier situation first, an endothermic reaction,where we add energy. This is as simple as heating up the container that the reactants are in, increasing the average speed of the molecules. Some of the molecules are moving slower than the average and some faster. If the molecules...

I think you need to consider why the bond is broken, that is, the bond will not spontaneously break. Say, for example, you have a chlorine molecule, it will not suddenly fly apart into its individual atoms, each gaining energy. That would violate the second law of thermodynamics. Imagine that...